I'm aware of the common way to prove that sqrt 2 is a real number but I'm just wondering would it suffice to prove it is a real number by proving it is an irrational number. As we know that irrational numbers are Real numbers.
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$\begingroup$Short answer: Yes!
Longer answer: Since the Irrationals are defined to be R\Q (The set of all real numbers such that the number is not a member of the rationals) then we can see R\Q is a subset of R. Then by definition of a subset if an element is in R\Q then it is also in R. So if you can prove that root 2 is irrational then you have proven it is a real number
$\endgroup$ $\begingroup$This question is not well-defined. The real numbers are typically constructed in one of three ways:
- Synethic axiomatization (i.e., complete ordered Archimedean field.)
- Dedekind Cuts
- Cauchy Sequences
Given each of these constructions, you would have a different proof that $\sqrt{2}$ is a real number. For example, given the Dedekind cut construction, it suffices to show $(A,B)$, such that $$\begin{align}A &= \left\{a \in \mathbb{Q}: a^2 < 2 \text{ or } a < 0 \right\} \\ B &= \left\{b \in \mathbb{Q} : b^2 > 2 \text{ and } b >0\right\} \end{align},$$ is a cut and that it is the square root of $2$. Under the Cauchy Sequence construction, we would typically work with a sequence such as $$\begin{align} A_0 &= 1 \\ A_{n+1} &= \frac{1}{2}(A_n + 2/A_n).\end{align}$$ At the end of the day, proving a number is a real number, depends on how you construct your real numbers.
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